What is the sum of 1370 cm, 1575 mm, and 8.63 m in meters

A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from

ivann1987 [24]

Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$ (1)

Where:

$I_{T}$ - Moment of inertia of the turntable, in kilogram-square meters.

$r$ - Distance of the block regarding the center of the turntable, in meters.

$m$ - Mass of the object, in kilograms.

$\omega_{o}$ - Initial angular speed of the turntable, in radians per second.

$\omega_{f}$ - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$ (2)

Where $M$ is the mass of the turntable, in kilograms.

If we know that $\omega_{o} \approx 7.330\,\frac{rad}{s}$ , $M = 1.5\,kg$ , $m = 0.54\,kg$ and $r = 0.1\,m$ , then the angular speed of the turntable after the event is:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$

$I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}$

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$

$\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}$

$\omega_{T} = 3.004\,\frac{rad}{s}$ ( $28.687\,\frac{rev}{min}$ )

The turntable's angular speed after the event is 28.687 revolutions per minute.

3 0

3 years ago

Are electric files lines due to an isolated positive point charge emanate radially away from the charge?

lara31 [8.8K]

Answer:

6

Explanation:

4 0

3 years ago

An 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its

nikdorinn [45]

Answer:

The average induced emf in the loop is 0.20 V

Explanation:

Given:

Radius of loop $r = \frac{d}{2} = 9.25 \times 10^{-2}$ m

Magnetic field $B = 1.5$ T

Change in time $\Delta t = 0.20$ sec

According to the faraday's law,

Induced emf is given by

$\epsilon = -\frac{\Delta \phi}{\Delta t}$

Where $\phi =$ magnetic flux

$\phi = BA\cos0$ ( here $\theta = 0$ )

Where $A = \pi r^{2}$

We neglect minus sign because it's shows lenz law

$\epsilon = \frac{B \pi r^{2} }{\Delta t}$

$\epsilon = \frac{1.5 \times 3.14 \times (9.25 \times 10^{-2} )^{2} }{0.20}$

$\epsilon = 0.20$ V

Therefore, the average induced emf in the loop is 0.20 V

4 0

4 years ago

the car starts from a stop to travel 1100 meters in 14 seconds. it is clocked at 65 m/s at point k. find its average speed and a

inysia [295]

Answer:

The average velocity of the car is, V = 74.04 m/s

Explanation:

Given data,

The initial velocity of the car, u = 0 m/s

The displacement of the ca, S = 1100 m

The time period of travel, t = 14 s

The velocity of the car at point k, v = 65 m/s

Using the II equation of motion,

S = ut + ½ at²

Substituting the given values,

1100 = 0 + ½ x a x 14²

a = 11.22 m/s²

Using the III equation of motion

v² = u² + 2 as

v = √(2as) (∵ u = 0)

Substituting,

v = √(2 x 11.22 x 1100)

= 157.11 m/s

The average speed of the car,

$V=\frac{0+65+157.11}{3}$

V = 74.04 m/s

Hence, the average velocity of the car is, V = 74.04 m/s

4 0

3 years ago

A light with a second-order bright band forms a diffraction angle of 30. 0°. The diffraction grating has 250. 0 lines per mm. Wh

Luden [163]

The distance between two successive troughs or crests is known as the wavelength. The wavelength of the light will be 1000 nm.

How do you define wavelength?

The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.

The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.

Diffraction angle= 30⁰

Diffraction grating per mm= 250

wavelength = ?

Mathematically the equation of bright band is given by

$\rm \lambda= \frac{sin\theta}{nN}$